add -1/2

3e83635a · Olivier Cots · 26b11cd9 · 3e83635a · 3e83635a
Commit 3e83635a authored 4 years ago by Olivier Cots
--- a/course/corrections/ex3_cor_multiple_shooting_bsb.ipynb
+++ b/course/corrections/ex3_cor_multiple_shooting_bsb.ipynb
@@ -365,7 +365,6 @@
   "outputs": [],
   "source": [
    "def dshoot(p0, dp0, e, tf):\n",
-    "#    ds = 0  ### TO COMPLETE\n",
    "    (xf, dxf), _ = f(t0, x0, (p0, dp0), tf, e)\n",
    "    ds = dxf\n",
    "    return ds\n",
@@ -894,7 +893,7 @@
    " \\begin{pmatrix}\n",
    "     x(t_1, t_0, x_0, p_0, u_-) \\\\\n",
    "     p(t_1, t_0, x_0, p_0, u_-) \\\\\n",
-    "     x(t_f, t_2, x_2, p_2, u_+)\n",
+    "     x(t_f, t_2, x_2, p_2, u_+) - 1/2\n",
    " \\end{pmatrix},\n",
    "$$\n",
    "\n",

 %% Cell type:markdown id: tags:
 # Exercice 3: Introduction to indirect multiple shooting: the Bang-Singular-Bang case on a turnpike problem
 %% Cell type:markdown id: tags:
 * Author: Olivier Cots
 * Date: March 2021
 ------
 * Back: [Exercice 2](../exercices/ex2_implement_simple_shooting.ipynb) -
 [Correction](../corrections/ex2_cor_implement_simple_shooting.ipynb)
 ------
 %% Cell type:markdown id: tags:
 ## I) Description of the optimal control problem
 %% Cell type:markdown id: tags:
 We consider the following optimal control problem:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} x^2(t) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 To this optimal control problem is associated the stationnary optimization problem
 $$
    \min_{(x, u)} \{~ x^2 ~ | ~  (x, u) \in \mathrm{R} \times [-1, 1],~ f(x,u) = u = 0\}.
 $$
 The static solution is thus $(x^*, u^*) = (0, 0)$. This solution may be seen as the static pair $(x, u)$ which minimizes the cost $J(u)$ under
 the constraint $u \in [-1, 1]$.
 It is well known that this problem is what we call a *turnpike* optimal control problem.
 Hence, if the final time $t_f$ is long enough the solution is of the following form:
 starting from $x(0)=1$, reach as fast as possible the static solution, stay at the static solution as long as possible before reaching
 the target $x(t_f)=1/2$. In this case, the optimal control would be
 $$
    u(t) = \left\{
    \begin{array}{lll}
        -1            & \text{if} & t \in [0, t_1],     \\[0.5em]
        \phantom{-}0  & \text{if} & t \in (t_1, t_2],   \\[0.5em]
        +1            & \text{if} & t \in (t_2, t_f],
    \end{array}
    \right.
 $$
 with $0 < t_1 < t_2 < t_f$. We say that the control is *Bang-Singular-Bang*. A Bang arc corresponds to $u \in \{-1, 1\}$ while a singular control corresponds to $u \in (-1, 1)$. Since the optimal control law is discontinuous, then to solve this optimal control problem by indirect methods and find the *switching times* $t_1$ and $t_2$, we need to implement what we call a *multiple shooting method*. In the next section we introduce a regularization technique to force the control to be in the set $(-1,1)$ and to be smooth. In this context, we will be able to implement a simple shooting method and determine the structure of the optimal control law. Thanks to the simple shooting method, we will have the structure of the optimal control law together with an approximation of the switching times that we will use as initial guess for the multiple shooting method that we present in the last section.
 <div class="alert alert-warning">
 **Main goal**
 Find the switching times $t_1$ and $t_2$ by multiple shooting.
 </div>
 Steps:
 1. Regularize the problem and solve the regularized problem by simple shooting.
 2. Determine the structure of the non-regularized optimal control problem, that is the structure Bang-Singular-Bang, and find a good approximation of the switching times and of the initial co-vector.
 3. Solve the non-regularized optimal control problem by multiple shooting.
 **_Remark 1._** See this [page](../lecture/lecture_simple_shooting.ipynb) for a general presentation of the simple shooting method.
 **_Remark 2._** In this particular example, the singular control does not depend on the costate $p$ since it is constant. This happens in low dimension. This could be taken into consideration to simplify the definition of the multiple shooting method. However, to stay general, we will not consider this particular property in this notebook.
 %% Cell type:markdown id: tags:
 ## II) Regularization and simple shooting
 %% Cell type:markdown id: tags:
 We make the following regularization:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} (x^2(t) - \varepsilon\ln(1-u^2(t))) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 Our goal is to determine the structure of the optimal control problem when $(\varepsilon, t_f) = (0, 2)$. The problem is simpler to solver when $\varepsilon$ is bigger and $t_f$ is smaller. It is also smooth whenever $\varepsilon>0$. Hence, we will start by solving the problem for $(\varepsilon, t_f) = (1, 1)$. In a second step, we will decrease the *penalization term* $\varepsilon$ and in a final step, we will increase the final time $t_f$ to the final value $2$.
 %% Cell type:markdown id: tags:
 ### Preliminaries
 %% Cell type:code id: tags:
 ``` python
 # import packages
 import nutopy as nt
 import nutopy.tools as tools
 import nutopy.ocp as ocp
 import numpy as np
 import matplotlib.pyplot as plt
 #plt.rcParams['figure.figsize'] = [10, 5]
 plt.rcParams['figure.dpi'] = 150
 ```
 %% Cell type:code id: tags:
 ``` python
 # Finite differences function for scalar functionnal
 # Return f'(x).dx
 def finite_diff(fun, x, dx, *args, **kwargs):
    v_eps = np.finfo(float).eps
    t = np.sqrt(v_eps) * np.sqrt(np.maximum(1.0, np.abs(x))) / np.sqrt(np.maximum(1.0, np.abs(dx)))
    j = (fun(x + t*dx, *args, **kwargs) - fun(x, *args, **kwargs)) / t
    return j
 ```
 %% Cell type:code id: tags:
 ``` python
 # Parameters
 t0        = 0.0
 x0        = np.array([1.0])
 xf_target = np.array([0.5])
 e_init    = 1.0
 e_final   = 0.002 #
 tf_init   = 1.0 # With this value the problem is simpler to solver since the trajectory stay
                # less time around the static solution
 tf_final  = 2.0
 ```
 %% Cell type:markdown id: tags:
 ### Maximized Hamiltonian and its derivatives
 %% Cell type:markdown id: tags:
 The pseudo-Hamiltonian is (in the normal case)
 $$
    H(x,p,u,\varepsilon) = pu - x^2 + \varepsilon ln(1-u^2).
 $$
 Note that we put the parameter $\varepsilon$ into the arguments of the pseudo-Hamiltonian since we will vary it.
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 1:_**
 Give the maximizing control $u[p, \varepsilon]$, that is the control in feedback form solution of the maximization condition.
 </div>
 %% Cell type:markdown id: tags:
 **Answer 1:** To complete here (double-click on the line to complete)
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 2:_**
 Complete the code of the maximizing control $u[p, \varepsilon]$ and its derivative with respect to $p$, that is $\frac{\partial u}{\partial p}[p, \varepsilon]$.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 2 to complete here
 # ----------------------------
 #
 # Control in feedback form u[p,e] and its partial derivative wrt. p.
 #
 @tools.vectorize(vvars=(1,))
 def ufun(p, e):
 #    u = 0  ### TO COMPLETE
    u = (-e + np.sqrt(e**2+p**2))/p
    return u
 def dufun(p, e):
 #    du = 0  ### TO COMPLETE
    s  = np.sqrt(e**2+p**2)
    du = 1.0/s + (e - s)/p**2
    return du
 ```
 %% Cell type:markdown id: tags:
 We give next the maximized Hamiltonian with its derivatives. This permits us to define the flow of the associated Hamiltonian vector field.
 %% Cell type:code id: tags:
 ``` python
 # Definition of the maximized Hamiltonian and its derivatives
 # The second derivative d2hfun is computed by finite differences for a part
 def dhfun(t, x, dx, p, dp, e):
    # dh = dh_x dx + dh_p dp
    u  = ufun(p, e)
    du = dufun(p, e)
    hd = (u+p*du+2.0*e*u*du/(u**2-1.0))*dp - 2.0*x*dx
    return hd
 def d2hfun(t, x, dx, d2x, p, dp, d2p, e):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    d2h_xx = -2.0*dx*d2x # dh_xx dx d2x
    dh_p   = lambda p: dhfun(t, x, 0.0, p, dp, e) # dh_px = 0 so we can put dx = 0
    d2h_pp = finite_diff(dh_p, p, d2p) # dh_pp dp d2p
    hdd    = d2h_xx + d2h_pp
    return hdd
 @tools.tensorize(dhfun, d2hfun, tvars=(2, 3))
 def hfun(t, x, p, e):
    u = ufun(p, e)
    h = p*u - x**2 + e*(np.log(1.0-u**2))
    return h
 h = ocp.Hamiltonian(hfun)   # The Hamiltonian object
 f = ocp.Flow(h)             # The flow associated to the Hamiltonian object is
                            # the exponential mapping with its derivative
                            # that can be used to define the Jacobian of the
                            # shooting function
 ```
 %% Cell type:markdown id: tags:
 ### Shooting function and its derivative
 %% Cell type:markdown id: tags:
 The shooting function is
 $$
    S(p_0, \varepsilon, t_f) = \pi_x(z(t_f, 1, p_0, \varepsilon)) - 1/2,
 $$
 where $z(t_f, x_0, p_0, \varepsilon)$ is the solution of the associated Hamiltonian system
 with the initial condition $z(0) = (x_0, p_0)$. Note that the Hamiltonian system depends on $\varepsilon$. We put $\varepsilon$ and $t_f$ into
 the arguments of the shooting function since we will vary them.
 <div class="alert alert-warning">
 **Procedure**
 First solve $S=0$ for $(\varepsilon, t_f) = (1,1)$ then decrease $\varepsilon$ to $0.01$, and finish by increasing $t_f$ to 2.
 </div>
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 3:_**
 Complete the code of the shooting function.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 3 to complete here
 # ----------------------------
 #
 # Definition of the shooting function and its partial derivative wrt. p0 against the vector dp0
 #
 def shoot(p0, e, tf):
 #    s = 0  ### TO COMPLETE: use the flow f, the parameters t0, x0 and xf_target
    xf, _ = f(t0, x0, p0, tf, e)  # We use the flow to get z(tf, x0, p0)
    s = xf - xf_target  # x(tf, x0, p0) - xf_target
    return s
 ```
 %% Cell type:code id: tags:
 ``` python
 def dshoot(p0, dp0, e, tf):
-#    ds = 0  ### TO COMPLETE
    (xf, dxf), _ = f(t0, x0, (p0, dp0), tf, e)
    ds = dxf
    return ds
 shoot = nt.tools.tensorize(dshoot, tvars=(1,))(shoot)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Function to plot the solution
 def plotSolution(p0, e, tf):
    N      = 200
    tspan  = list(np.linspace(t0, tf, N+1))
    xf, pf = f(t0, x0, p0, tspan, e)
    u      = ufun(pf, e)
    fig = plt.figure()
    ax  = fig.add_subplot(511); ax.plot(tspan, xf); ax.set_xlabel('t'); ax.set_ylabel('$x$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(513); ax.plot(tspan, pf); ax.set_xlabel('t'); ax.set_ylabel('$p$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(515); ax.plot(tspan,  u); ax.set_xlabel('t'); ax.set_ylabel('$u$'); ax.axhline(0, color='k')
 ```
 %% Cell type:markdown id: tags:
 ### Resolution of the regularized problem
 %% Cell type:code id: tags:
 ``` python
 # Shooting for (tf, e) = (tf_init, e_init)
 p0_guess = np.array([0.1])
 nlefun   = lambda p0: shoot(p0, e_init, tf_init)
 sol_nle  = nt.nle.solve(nlefun, p0_guess, df=nlefun)
 ```
 %% Output
         Calls  |f(x)|                 |x|
             1  9.225527956247914e-01  1.000000000000000e-01
             2  1.328510039876105e-01  2.933843238918098e+00
             3  7.295873671084141e-02  2.551952326284172e+00
             4  3.048709252432280e-02  2.086745717585143e+00
             5  4.420525056492819e-03  2.223849330434452e+00
             6  2.283580968987509e-04  2.206487218725369e+00
             7  1.831090794324197e-06  2.205541459926580e+00
             8  7.509122212923103e-10  2.205548983174077e+00
             9  2.775557561562891e-15  2.205548980090132e+00
     Results of the nle solver method:
     xsol    =  [-2.20554898]
     f(xsol) =  [-2.77555756e-15]
     nfev    =  9
     njev    =  1
     status  =  1
     success =  True
     Successfully completed: relative error between two consecutive iterates is at most TolX.
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_init, e_init)
 plotSolution(sol_nle.x, e_init, tf_init)
 ```
 %% Output
 %% Cell type:code id: tags:
 ``` python
 # Definition of the homotopic function and its first order derivative
 # This function is used to solve S=0 for different values of e=epsilon and tf.
 def dhomfun(p0, dp0, e, de, tf, dtf):
    #
    (xf, dxf), (pf, dpf) = f(t0, x0, (p0, dp0), tf, e)
    #
    s     = xf - xf_target
    #
    ds_p0 = dxf
    ds_tf = ufun(pf, e) * dtf # dS_tf dtf = u dtf
    #
    fun = lambda e: float(f(t0, x0, p0, tf, e)[0])
    ds_e = finite_diff(fun, e, de) # dS_e de
    #
    ds    = ds_p0 + ds_e + ds_tf
    return s, ds
 @tools.tensorize(dhomfun, tvars=(1, 2, 3), full=True)
 def homfun(p0, e, tf):
    xf, pf = f(t0, x0, p0, tf, e)  # We use the flow to get z(tf, x0, p0, e)
    s = xf - xf_target             # x(tf, x0, p0) - xf_target
    return s
 ```
 %% Cell type:code id: tags:
 ``` python
 # Making the penalization smaller: homotopy on e
 p0         = sol_nle.x
 sol_path_e = nt.path.solve(homfun, p0, e_init, e_final, args=tf_init, df=homfun)
 ```
 %% Output
         Calls  |f(x,pars)|     |x|                Homotopic param    Arclength s     det A(s)        Dot product
             1  2.22044605e-16  2.20554898009e+00  1.00000000000e+00  0.00000000e+00 -3.99453543e-01  0.00000000e+00
             2  2.22044605e-16  2.19703976141e+00  9.93456169329e-01  1.07344549e-02 -4.02221666e-01  9.99999991e-01
             3  3.33066907e-16  2.17562949518e+00  9.76983031043e-01  3.77485955e-02 -4.09360819e-01  9.99999942e-01
             4  2.22044605e-16  2.13582370908e+00  9.46324313715e-01  8.79925768e-02 -4.23336260e-01  9.99999773e-01
             5  0.00000000e+00  2.04380511707e+00  8.75274200125e-01  2.04248946e-01 -4.59644793e-01  9.99998415e-01
             6  1.11022302e-16  1.92916643462e+00  7.86347624598e-01  3.49335047e-01 -5.14727915e-01  9.99996190e-01
             7  2.22044605e-16  1.79444755857e+00  6.81092176321e-01  5.20296827e-01 -5.99279037e-01  9.99990735e-01
             8  1.11022302e-16  1.66966322809e+00  5.82619647458e-01  6.79256037e-01 -7.06991902e-01  9.99985113e-01
             9  1.11022302e-16  1.56620772374e+00  5.00008221016e-01  8.11648417e-01 -8.30695653e-01  9.99981810e-01
            10  3.33066907e-16  1.46593779252e+00  4.18824463927e-01  9.40663687e-01 -9.99327396e-01  9.99971770e-01
            11  0.00000000e+00  1.37678846132e+00  3.45480162266e-01  1.05610658e+00 -1.21575495e+00  9.99967683e-01
            12  5.55111512e-17  1.28960342151e+00  2.72548876780e-01  1.16977396e+00 -1.52984851e+00  9.99968620e-01
            13  1.11022302e-16  1.23722039409e+00  2.28228170219e-01  1.23839109e+00 -1.79572280e+00  9.99995132e-01
            14  7.77156117e-16  1.19422849087e+00  1.91735175890e-01  1.29478295e+00 -2.07744292e+00  9.99999997e-01
            15  8.88178420e-16  1.15761257447e+00  1.60763287717e-01  1.34274112e+00 -2.37832445e+00  9.99993905e-01
            16  8.88178420e-16  1.12621332813e+00  1.34479524463e-01  1.38368932e+00 -2.69537295e+00  9.99976683e-01
            17  6.66133815e-16  1.09882012933e+00  1.11936744078e-01  1.41916572e+00 -3.03058070e+00  9.99950610e-01
            18  9.43689571e-16  1.07488689408e+00  9.26897526120e-02  1.44987824e+00 -3.38245451e+00  9.99920746e-01
            19  3.10862447e-15  1.05162281210e+00  7.45308292960e-02  1.47939067e+00 -3.79520044e+00  9.99862606e-01
            20  4.44089210e-16  1.03348997369e+00  6.08568585866e-02  1.50210166e+00 -4.18199975e+00  9.99863622e-01
            21  7.77156117e-16  1.01743621699e+00  4.91774488730e-02  1.52195467e+00 -4.58879415e+00  9.99842046e-01
            22  2.88657986e-15  1.00351364714e+00  3.94278240330e-02  1.53895177e+00 -5.00712133e+00  9.99833210e-01
            23  1.11022302e-15  9.91189285569e-01  3.11338118442e-02  1.55380731e+00 -5.44570207e+00  9.99823057e-01
            24  7.21644966e-16  9.80356390822e-01  2.41377538128e-02  1.56670310e+00 -5.90269325e+00  9.99819071e-01
            25  1.11022302e-16  9.70835442114e-01  1.82431398575e-02  1.57790126e+00 -6.37956456e+00  9.99817622e-01
            26  8.99280650e-15  9.62692462871e-01  1.34115980809e-02  1.58736986e+00 -6.86424447e+00  9.99827368e-01
            27  4.77395901e-15  9.55675551884e-01  9.42082903289e-03  1.59544235e+00 -7.36122521e+00  9.99834372e-01
            28  1.32116540e-14  9.49853651601e-01  6.24486269889e-03  1.60207427e+00 -7.85282352e+00  9.99851794e-01
            29  4.19664303e-14  9.45389069838e-01  3.90225115046e-03  1.60711617e+00 -8.30049315e+00  9.99886223e-01
            30  1.88737914e-15  9.41931793716e-01  2.15059352872e-03  1.61099191e+00 -8.71023410e+00  9.99909978e-01
            31  6.66133815e-16  9.41628939496e-01  2.00000000000e-03  1.61133013e+00 -8.74973421e+00  9.99999182e-01
     Results of the path solver method:
     xf            =  [-0.94162894]
     parsf         =  0.002
     |f(xf,parsf)| =  6.661338147750939e-16
     steps         =  31
     status        =  1
     success       =  True
     Homotopy successfully completed.
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_init, e_final)
 plotSolution(sol_path_e.xf, e_final, tf_init)
 ```
 %% Output
 %% Cell type:code id: tags:
 ``` python
 # Making the time bigger: homotopy on tf
 p0          = sol_path_e.xf                       # sol is coming from last homotopy
 pathfun     = lambda p0, tf, e: homfun(p0, e, tf) # invert order of arguments
 sol_path_tf = nt.path.solve(pathfun, p0, tf_init, tf_final, args=e_final, df=pathfun)
 ```
 %% Output
         Calls  |f(x,pars)|     |x|                Homotopic param    Arclength s     det A(s)        Dot product
             1  2.22044605e-16  9.41628939496e-01  1.00000000000e+00  0.00000000e+00  4.01455484e+00  0.00000000e+00
             2  8.88178420e-16  9.44075168781e-01  1.00970881514e+00  1.00122571e-02  4.08611419e+00  9.99990245e-01
             3  6.66133815e-16  9.52110673458e-01  1.04297202177e+00  4.42326128e-02  4.35312513e+00  9.99884749e-01
             4  8.88178420e-16  9.61556842283e-01  1.08526416490e+00  8.75675217e-02  4.75042396e+00  9.99811995e-01
             5  5.55111512e-17  9.71896630561e-01  1.13674411999e+00  1.40076814e-01  5.34799409e+00  9.99719724e-01
             6  5.55111512e-16  9.81245191618e-01  1.18988053242e+00  1.94030671e-01  6.14816857e+00  9.99701750e-01
             7  2.22044605e-15  9.89208695747e-01  1.24261341184e+00  2.47362758e-01  7.21580053e+00  9.99709964e-01
             8  7.21644966e-16  9.95584144595e-01  1.29272773597e+00  2.97882063e-01  8.62227225e+00  9.99745289e-01
             9  1.22124533e-15  1.00051725716e+00  1.33956581293e+00  3.44980041e-01  1.04922584e+01  9.99787850e-01
            10  1.88737914e-15  1.00422155325e+00  1.38281120954e+00  3.88384407e-01  1.30075715e+01  9.99831827e-01
            11  2.33146835e-15  1.00693858544e+00  1.42255002654e+00  4.28216425e-01  1.64500266e+01  9.99872114e-01
            12  4.88498131e-15  1.00888895880e+00  1.45901756959e+00  4.64736370e-01  2.12529132e+01  9.99906881e-01
            13  1.22124533e-14  1.01026195185e+00  1.49256266733e+00  4.98309736e-01  2.80968027e+01  9.99935247e-01
            14  4.94049246e-15  1.01121117641e+00  1.52358925456e+00  5.29350951e-01  3.80663163e+01  9.99957193e-01
            15  6.59472477e-14  1.01185675334e+00  1.55253698600e+00  5.58305944e-01  5.29258454e+01  9.99973207e-01
            16  8.80406859e-14  1.01228937436e+00  1.57986254022e+00  5.85634959e-01  7.56112643e+01  9.99984175e-01
            17  1.68531855e-13  1.01257542227e+00  1.60602989124e+00  6.11803893e-01  1.11142150e+02  9.99991198e-01
            18  2.80664381e-13  1.01276214622e+00  1.63150346762e+00  6.37278163e-01  1.68372602e+02  9.99995395e-01
            19  2.86659585e-13  1.01288241210e+00  1.65674705661e+00  6.62522043e-01  2.63490945e+02  9.99997737e-01
            20  2.96984659e-13  1.01295868532e+00  1.68222816553e+00  6.88003269e-01  4.27333674e+02  9.99998957e-01
            21  1.11244347e-13  1.01300613907e+00  1.70842815072e+00  7.14203298e-01  7.21468310e+02  9.99999552e-01
            22  4.27813340e-12  1.01303494245e+00  1.73585852757e+00  7.41633690e-01  1.27569212e+03  9.99999821e-01
            23  2.57732169e-11  1.01305187229e+00  1.76508420787e+00  7.70859375e-01  2.38160109e+03  9.99999935e-01
            24  1.20242538e-10  1.01306141530e+00  1.79675602397e+00  8.02531193e-01  4.74536748e+03  9.99999978e-01
            25  6.23208152e-10  1.01306651002e+00  1.83165732709e+00  8.37432497e-01  1.02379807e+04  9.99999994e-01
            26  3.66588537e-09  1.01306904469e+00  1.87077329303e+00  8.76548463e-01  2.43881436e+04  9.99999998e-01
            27  2.69244553e-08  1.01307019491e+00  1.91539837246e+00  9.21173542e-01  6.58870660e+04  1.00000000e+00
            28  2.69732114e-07  1.01307065731e+00  1.96730883161e+00  9.73084001e-01  2.09516626e+05  1.00000000e+00
            29  2.15117862e-06  1.01307076680e+00  2.00000000000e+00  1.00577516e+00  4.36713283e+05  1.00000000e+00
     Results of the path solver method:
     xf            =  [-1.01307077]
     parsf         =  2.0
     |f(xf,parsf)| =  2.1511786191807936e-06
     steps         =  29
     status        =  1
     success       =  True
     Homotopy successfully completed.
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_final, e_final)
 plotSolution(sol_path_tf.xf, e_final, tf_final)
 ```
 %% Output
 %% Cell type:markdown id: tags:
 ## III) Resolution of the optimal control problem by multiple shooting
 %% Cell type:markdown id: tags:
 We come back to the original optimal control problem:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} x^2(t) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 We have determined that the optimal control follows the strategy:
 $$
    u(t) = \left\{
    \begin{array}{lll}
        -1            & \text{if} & t \in [0, t_1],     \\[0.5em]
        \phantom{-}0  & \text{if} & t \in (t_1, t_2],   \\[0.5em]
        +1            & \text{if} & t \in (t_2, t_f],
    \end{array}
    \right.
 $$
 with $0 < t_1 < t_2 < t_f=2$.
 <div class="alert alert-warning">
 **Goal**
 The goal is to find the values of the switching times $t_1$ and $t_2$ together with the initial covector $p_0$ (see Remark 2).
 </div>
 %% Cell type:markdown id: tags:
 ### Maximized Hamiltonian and its derivatives
 %% Cell type:markdown id: tags:
 We define first the three control laws $u \equiv \{-1, 0, 1\}$.
 %% Cell type:code id: tags:
 ``` python
 # Controls in feedback form
 @tools.vectorize(vvars=(1, 2, 3))
 def uplus(t, x, p):
    u = +1.0
    return u
 @tools.vectorize(vvars=(1, 2, 3))
 def uminus(t, x, p):
    u = -1.0
    return u
 @tools.vectorize(vvars=(1, 2, 3))
 def using(t, x, p):
    u = 0.0
    return u
 ```
 %% Cell type:markdown id: tags:
 The pseudo-Hamiltonian is
 $$
    H(x,p,u) = pu - x^2.
 $$
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 4:_**
 Complete the code of the Hamiltonian for $u \equiv +1$, with its derivatives.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 4 to complete here
 # ----------------------------
 #
 # Definition of the Hamiltonian and its derivatives for u = 1
 #
 def dhfunplus(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
 #    hd = 0 ### TO COMPLETE: use uplus
    u  = uplus(t, x, p)
    hd = u*dp - 2.0*x*dx
    return hd
 def d2hfunplus(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
 #    hdd = 0 ### TO COMPLETE
    hdd    = -2.0 * d2x * dx
    return hdd
 @tools.tensorize(dhfunplus, d2hfunplus, tvars=(2, 3))
 def hfunplus(t, x, p):
 #    h = 0 ### TO COMPLETE: use uplus
    u = uplus(t, x, p)
    h = p*u - x**2
    return h
 hplus = ocp.Hamiltonian(hfunplus)
 fplus = ocp.Flow(hplus)
 ```
 %% Cell type:markdown id: tags:
 We give the Hamiltonians for $u=-1$ and $u=0$ with their derivatives.
 %% Cell type:code id: tags:
 ``` python
 # Definition of the Hamiltonian and its derivatives for u = -1
 def dhfunminus(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
    u  = uminus(t, x, p)
    hd = u*dp - 2.0*x*dx
    return hd
 def d2hfunminus(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    hdd    = -2.0 * d2x * dx
    return hdd
 @tools.tensorize(dhfunminus, d2hfunminus, tvars=(2, 3))
 def hfunminus(t, x, p):
    u = uminus(t, x, p)
    h = p*u - x**2
    return h
 hminus = ocp.Hamiltonian(hfunminus)
 fminus = ocp.Flow(hminus)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Definition of the Hamiltonian and its derivatives for u = 0
 def dhfunsing(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
    u  = using(t, x, p)
    hd = u*dp - 2.0*x*dx
    return hd
 def d2hfunsing(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    hdd    = -2.0 * d2x * dx
    return hdd
 @tools.tensorize(dhfunsing, d2hfunsing, tvars=(2, 3))
 def hfunsing(t, x, p):
    u = using(t, x, p)
    h = p*u - x**2
    return h
 hsing = ocp.Hamiltonian(hfunsing)
 fsing = ocp.Flow(hsing)
 ```
 %% Cell type:markdown id: tags:
 ### Shooting function
 %% Cell type:markdown id: tags:
 The multiple shooting function is given by
 $$
 S(p_0, t_1, t_2) :=
 \begin{pmatrix}
     x(t_1, t_0, x_0, p_0, u_-) \\
     p(t_1, t_0, x_0, p_0, u_-) \\
-     x(t_f, t_2, x_2, p_2, u_+)
+     x(t_f, t_2, x_2, p_2, u_+) - 1/2
 \end{pmatrix},
 $$
 where $z_2 := (x_2, p_2) = z(t_2, t_1, x_1, p_1, u_0)$, $z_1 := (x_1, p_1) = z(t_1, t_0, x_0, p_0, u_-)$ and where z(t, s, a, b, u) is the solution at time $t$ of the Hamiltonian system associated to the control u starting at time $s$ at the initial condition $z(s) = (a,b)$.
 We have introduced the notation $u_-$ for $u\equiv -1$, $u_0$ for $u\equiv 0$ and $u_+$ for $u\equiv +1$.
 **_Remark:_** We know that $(x_2, p_2)=(0,0)$.
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 5:_**
 Complete the code of the multiple shooting function.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 5 to complete here
 # ----------------------------
 #
 # Multiple shooting function
 #
 tf = tf_final # we set the final time to the value tf_final
 def shoot_multiple(y):
    p0 = y[0]
    t1 = y[1]
    t2 = y[2]
 #    s = np.zeros([3]) ### TO COMPLETE: use fminus, fsin, fplus, t0, t1, t2, tf, x0, xf_target
    x1, p1 = fminus(t0, x0, p0, t1)  # on [ 0, t1]
    x2, p2 =  (np.array([0.0]), np.array([0.0])) # fsing(t1, x1, p1, t2)  # on [t1, t2]
    xf, pf =  fplus(t2, x2, p2, tf)  # on [t2, tf]
    s = np.zeros([3])
    s[0] = x1
    s[1] = p1
    s[2] = xf - xf_target  # x(tf, x0, p0) - xf_target
    return s
 ```
 %% Cell type:markdown id: tags:
 ### Resolution of the shooting function
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 6:_**
 Give initial guesses for the times $t_1$ and $t_2$ according to the solution of the regularized problem.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 6 to complete here
 # ----------------------------
 #
 # Initial guess for the Newton solver
 t1_guess = 0.9 # to update
 t2_guess = 1.6 # to update
 p0_guess = sol_path_tf.xf # from previous homotopy
 ```
 %% Cell type:code id: tags:
 ``` python
 # Resolution of the shooting function
 y_guess  = np.array([float(p0_guess), t1_guess, t2_guess])
 sol_nle_mul  = nt.nle.solve(shoot_multiple, y_guess)
 ```
 %% Output
         Calls  |f(x)|                 |x|
             1  1.432908241330024e-01  2.096738509815576e+00
             2  9.999998053989939e-03  2.066422027958497e+00
             3  1.506856008950395e-05  2.061545503561054e+00
             4  1.103251344504288e-11  2.061552812803479e+00
             5  3.854837484910665e-16  2.061552812808831e+00
     Results of the nle solver method:
     xsol    =  [-1.   1.   1.5]
     f(xsol) =  [ 1.38777878e-17  3.47378376e-16 -1.66533454e-16]
     nfev    =  5
     njev    =  1
     status  =  1
     success =  True
     Successfully completed: relative error between two consecutive iterates is at most TolX.
 %% Cell type:code id: tags:
 ``` python
 # function to plot solution
 def plotSolutionBSB(p0, t1, t2, tf):
    N      = 20
    tspan1  = list(np.linspace(t0, t1, N+1))
    tspan2  = list(np.linspace(t1, t2, N+1))
    tspanf  = list(np.linspace(t2, tf, N+1))
    x1, p1 = fminus(t0, x0, p0, tspan1)  # on [ 0, t1]
    x2, p2 =  fsing(t1, x1[-1], p1[-1], tspan2)  # on [t1, t2]
    xf, pf =  fplus(t2, x2[-1], p2[-1], tspanf)  # on [t2, tf]
    u1     = uminus(tspan1, x1, p1)
    u2     =  using(tspan2, x2, p2)
    uf     =  uplus(tspanf, xf, pf)
    fig = plt.figure()
    ax  = fig.add_subplot(511); ax.plot(tspan1, x1); ax.plot(tspan2, x2); ax.plot(tspanf, xf);
    ax.set_xlabel('t'); ax.set_ylabel('$x$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(513);  ax.plot(tspan1, p1); ax.plot(tspan2, p2); ax.plot(tspanf, pf);
    ax.set_xlabel('t'); ax.set_ylabel('$p$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(515);  ax.plot(tspan1, u1); ax.plot(tspan2, u2); ax.plot(tspanf, uf);
    ax.set_xlabel('t'); ax.set_ylabel('$u$'); ax.axhline(0, color='k')
 ```
 %% Cell type:code id: tags:
 ``` python
 # plot solution
 p0 = sol_nle_mul.x[0]
 t1 = sol_nle_mul.x[1]
 t2 = sol_nle_mul.x[2]
 plotSolutionBSB(p0, t1, t2, tf)
 ```
 %% Output

--- a/course/exercices/ex3_multiple_shooting_bsb.ipynb
+++ b/course/exercices/ex3_multiple_shooting_bsb.ipynb
@@ -708,7 +708,7 @@
    " \\begin{pmatrix}\n",
    "     x(t_1, t_0, x_0, p_0, u_-) \\\\\n",
    "     p(t_1, t_0, x_0, p_0, u_-) \\\\\n",
-    "     x(t_f, t_2, x_2, p_2, u_+)\n",
+    "     x(t_f, t_2, x_2, p_2, u_+) - 1/2\n",
    " \\end{pmatrix},\n",
    "$$\n",
    "\n",

 %% Cell type:markdown id: tags:
 # Exercice 3: Introduction to indirect multiple shooting: the Bang-Singular-Bang case on a turnpike problem
 %% Cell type:markdown id: tags:
 * Author: Olivier Cots
 * Date: March 2021
 ------
 * Back: [Exercice 2](../exercices/ex2_implement_simple_shooting.ipynb) -
 [Correction](../corrections/ex2_cor_implement_simple_shooting.ipynb)
 ------
 %% Cell type:markdown id: tags:
 ## I) Description of the optimal control problem
 %% Cell type:markdown id: tags:
 We consider the following optimal control problem:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} x^2(t) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 To this optimal control problem is associated the stationnary optimization problem
 $$
    \min_{(x, u)} \{~ x^2 ~ | ~  (x, u) \in \mathrm{R} \times [-1, 1],~ f(x,u) = u = 0\}.
 $$
 The static solution is thus $(x^*, u^*) = (0, 0)$. This solution may be seen as the static pair $(x, u)$ which minimizes the cost $J(u)$ under
 the constraint $u \in [-1, 1]$.
 It is well known that this problem is what we call a *turnpike* optimal control problem.
 Hence, if the final time $t_f$ is long enough the solution is of the following form:
 starting from $x(0)=1$, reach as fast as possible the static solution, stay at the static solution as long as possible before reaching
 the target $x(t_f)=1/2$. In this case, the optimal control would be
 $$
    u(t) = \left\{
    \begin{array}{lll}
        -1            & \text{if} & t \in [0, t_1],     \\[0.5em]
        \phantom{-}0  & \text{if} & t \in (t_1, t_2],   \\[0.5em]
        +1            & \text{if} & t \in (t_2, t_f],
    \end{array}
    \right.
 $$
 with $0 < t_1 < t_2 < t_f$. We say that the control is *Bang-Singular-Bang*. A Bang arc corresponds to $u \in \{-1, 1\}$ while a singular control corresponds to $u \in (-1, 1)$. Since the optimal control law is discontinuous, then to solve this optimal control problem by indirect methods and find the *switching times* $t_1$ and $t_2$, we need to implement what we call a *multiple shooting method*. In the next section we introduce a regularization technique to force the control to be in the set $(-1,1)$ and to be smooth. In this context, we will be able to implement a simple shooting method and determine the structure of the optimal control law. Thanks to the simple shooting method, we will have the structure of the optimal control law together with an approximation of the switching times that we will use as initial guess for the multiple shooting method that we present in the last section.
 <div class="alert alert-warning">
 **Main goal**
 Find the switching times $t_1$ and $t_2$ by multiple shooting.
 </div>
 Steps:
 1. Regularize the problem and solve the regularized problem by simple shooting.
 2. Determine the structure of the non-regularized optimal control problem, that is the structure Bang-Singular-Bang, and find a good approximation of the switching times and of the initial co-vector.
 3. Solve the non-regularized optimal control problem by multiple shooting.
 **_Remark 1._** See this [page](../lecture/lecture_simple_shooting.ipynb) for a general presentation of the simple shooting method.
 **_Remark 2._** In this particular example, the singular control does not depend on the costate $p$ since it is constant. This happens in low dimension. This could be taken into consideration to simplify the definition of the multiple shooting method. However, to stay general, we will not consider this particular property in this notebook.
 %% Cell type:markdown id: tags:
 ## II) Regularization and simple shooting
 %% Cell type:markdown id: tags:
 We make the following regularization:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} (x^2(t) - \varepsilon\ln(1-u^2(t))) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 Our goal is to determine the structure of the optimal control problem when $(\varepsilon, t_f) = (0, 2)$. The problem is simpler to solver when $\varepsilon$ is bigger and $t_f$ is smaller. It is also smooth whenever $\varepsilon>0$. Hence, we will start by solving the problem for $(\varepsilon, t_f) = (1, 1)$. In a second step, we will decrease the *penalization term* $\varepsilon$ and in a final step, we will increase the final time $t_f$ to the final value $2$.
 %% Cell type:markdown id: tags:
 ### Preliminaries
 %% Cell type:code id: tags:
 ``` python
 # import packages
 import nutopy as nt
 import nutopy.tools as tools
 import nutopy.ocp as ocp
 import numpy as np
 import matplotlib.pyplot as plt
 #plt.rcParams['figure.figsize'] = [10, 5]
 plt.rcParams['figure.dpi'] = 150
 ```
 %% Cell type:code id: tags:
 ``` python
 # Finite differences function for scalar functionnal
 # Return f'(x).dx
 def finite_diff(fun, x, dx, *args, **kwargs):
    v_eps = np.finfo(float).eps
    t = np.sqrt(v_eps) * np.sqrt(np.maximum(1.0, np.abs(x))) / np.sqrt(np.maximum(1.0, np.abs(dx)))
    j = (fun(x + t*dx, *args, **kwargs) - fun(x, *args, **kwargs)) / t
    return j
 ```
 %% Cell type:code id: tags:
 ``` python
 # Parameters
 t0        = 0.0
 x0        = np.array([1.0])
 xf_target = np.array([0.5])
 e_init    = 1.0
 e_final   = 0.002 #
 tf_init   = 1.0 # With this value the problem is simpler to solver since the trajectory stay
                # less time around the static solution
 tf_final  = 2.0
 ```
 %% Cell type:markdown id: tags:
 ### Maximized Hamiltonian and its derivatives
 %% Cell type:markdown id: tags:
 The pseudo-Hamiltonian is (in the normal case)
 $$
    H(x,p,u,\varepsilon) = pu - x^2 + \varepsilon ln(1-u^2).
 $$
 Note that we put the parameter $\varepsilon$ into the arguments of the pseudo-Hamiltonian since we will vary it.
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 1:_**
 Give the maximizing control $u[p, \varepsilon]$, that is the control in feedback form solution of the maximization condition.
 </div>
 %% Cell type:markdown id: tags:
 **Answer 1:** To complete here (double-click on the line to complete)
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 2:_**
 Complete the code of the maximizing control $u[p, \varepsilon]$ and its derivative with respect to $p$, that is $\frac{\partial u}{\partial p}[p, \varepsilon]$.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 2 to complete here
 # ----------------------------
 #
 # Control in feedback form u[p,e] and its partial derivative wrt. p.
 #
 @tools.vectorize(vvars=(1,))
 def ufun(p, e):
    u = 0  ### TO COMPLETE
    return u
 def dufun(p, e):
    du = 0  ### TO COMPLETE
    return du
 ```
 %% Cell type:markdown id: tags:
 We give next the maximized Hamiltonian with its derivatives. This permits us to define the flow of the associated Hamiltonian vector field.
 %% Cell type:code id: tags:
 ``` python
 # Definition of the maximized Hamiltonian and its derivatives
 # The second derivative d2hfun is computed by finite differences for a part
 def dhfun(t, x, dx, p, dp, e):
    # dh = dh_x dx + dh_p dp
    u  = ufun(p, e)
    du = dufun(p, e)
    hd = (u+p*du+2.0*e*u*du/(u**2-1.0))*dp - 2.0*x*dx
    return hd
 def d2hfun(t, x, dx, d2x, p, dp, d2p, e):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    d2h_xx = -2.0*dx*d2x # dh_xx dx d2x
    dh_p   = lambda p: dhfun(t, x, 0.0, p, dp, e) # dh_px = 0 so we can put dx = 0
    d2h_pp = finite_diff(dh_p, p, d2p) # dh_pp dp d2p
    hdd    = d2h_xx + d2h_pp
    return hdd
 @tools.tensorize(dhfun, d2hfun, tvars=(2, 3))
 def hfun(t, x, p, e):
    u = ufun(p, e)
    h = p*u - x**2 + e*(np.log(1.0-u**2))
    return h
 h = ocp.Hamiltonian(hfun)   # The Hamiltonian object
 f = ocp.Flow(h)             # The flow associated to the Hamiltonian object is
                            # the exponential mapping with its derivative
                            # that can be used to define the Jacobian of the
                            # shooting function
 ```
 %% Cell type:markdown id: tags:
 ### Shooting function and its derivative
 %% Cell type:markdown id: tags:
 The shooting function is
 $$
    S(p_0, \varepsilon, t_f) = \pi_x(z(t_f, 1, p_0, \varepsilon)) - 1/2,
 $$
 where $z(t_f, x_0, p_0, \varepsilon)$ is the solution of the associated Hamiltonian system
 with the initial condition $z(0) = (x_0, p_0)$. Note that the Hamiltonian system depends on $\varepsilon$. We put $\varepsilon$ and $t_f$ into
 the arguments of the shooting function since we will vary them.
 <div class="alert alert-warning">
 **Procedure**
 First solve $S=0$ for $(\varepsilon, t_f) = (1,1)$ then decrease $\varepsilon$ to $0.01$, and finish by increasing $t_f$ to 2.
 </div>
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 3:_**
 Complete the code of the shooting function.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 3 to complete here
 # ----------------------------
 #
 # Definition of the shooting function and its partial derivative wrt. p0 against the vector dp0
 #
 def shoot(p0, e, tf):
    s = 0  ### TO COMPLETE: use the flow f, the parameters t0, x0 and xf_target
    return s
 ```
 %% Cell type:code id: tags:
 ``` python
 def dshoot(p0, dp0, e, tf):
    (xf, dxf), _ = f(t0, x0, (p0, dp0), tf, e)
    ds = dxf
    return ds
 shoot = nt.tools.tensorize(dshoot, tvars=(1,))(shoot)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Function to plot the solution
 def plotSolution(p0, e, tf):
    N      = 200
    tspan  = list(np.linspace(t0, tf, N+1))
    xf, pf = f(t0, x0, p0, tspan, e)
    u      = ufun(pf, e)
    fig = plt.figure()
    ax  = fig.add_subplot(511); ax.plot(tspan, xf); ax.set_xlabel('t'); ax.set_ylabel('$x$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(513); ax.plot(tspan, pf); ax.set_xlabel('t'); ax.set_ylabel('$p$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(515); ax.plot(tspan,  u); ax.set_xlabel('t'); ax.set_ylabel('$u$'); ax.axhline(0, color='k')
 ```
 %% Cell type:markdown id: tags:
 ### Resolution of the regularized problem
 %% Cell type:code id: tags:
 ``` python
 # Shooting for (tf, e) = (tf_init, e_init)
 p0_guess = np.array([0.1])
 nlefun   = lambda p0: shoot(p0, e_init, tf_init)
 sol_nle  = nt.nle.solve(nlefun, p0_guess, df=nlefun)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_init, e_init)
 plotSolution(sol_nle.x, e_init, tf_init)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Definition of the homotopic function and its first order derivative
 # This function is used to solve S=0 for different values of e=epsilon and tf.
 def dhomfun(p0, dp0, e, de, tf, dtf):
    #
    (xf, dxf), (pf, dpf) = f(t0, x0, (p0, dp0), tf, e)
    #
    s     = xf - xf_target
    #
    ds_p0 = dxf
    ds_tf = ufun(pf, e) * dtf # dS_tf dtf = u dtf
    #
    fun = lambda e: float(f(t0, x0, p0, tf, e)[0])
    ds_e = finite_diff(fun, e, de) # dS_e de
    #
    ds    = ds_p0 + ds_e + ds_tf
    return s, ds
 @tools.tensorize(dhomfun, tvars=(1, 2, 3), full=True)
 def homfun(p0, e, tf):
    xf, pf = f(t0, x0, p0, tf, e)  # We use the flow to get z(tf, x0, p0, e)
    s = xf - xf_target             # x(tf, x0, p0) - xf_target
    return s
 ```
 %% Cell type:code id: tags:
 ``` python
 # Making the penalization smaller: homotopy on e
 p0         = sol_nle.x
 sol_path_e = nt.path.solve(homfun, p0, e_init, e_final, args=tf_init, df=homfun)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_init, e_final)
 plotSolution(sol_path_e.xf, e_final, tf_init)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Making the time bigger: homotopy on tf
 p0          = sol_path_e.xf                       # sol is coming from last homotopy
 pathfun     = lambda p0, tf, e: homfun(p0, e, tf) # invert order of arguments
 sol_path_tf = nt.path.solve(pathfun, p0, tf_init, tf_final, args=e_final, df=pathfun)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Plot solution for (tf, e) = (tf_final, e_final)
 plotSolution(sol_path_tf.xf, e_final, tf_final)
 ```
 %% Cell type:markdown id: tags:
 ## III) Resolution of the optimal control problem by multiple shooting
 %% Cell type:markdown id: tags:
 We come back to the original optimal control problem:
 $$
    \left\{
    \begin{array}{l}
        \displaystyle J(u)  := \displaystyle \int_0^{t_f} x^2(t) \, \mathrm{d}t \longrightarrow \min \\[1.0em]
        \dot{x}(t) = f(x(t), u(t)) := \displaystyle u(t), \quad  |u(t)| \le 1, \quad t \in [0, t_f] \text{ a.e.},    \\[1.0em]
        x(0) = 1, \quad x(t_f) = 1/2.
    \end{array}
    \right.
 $$
 We have determined that the optimal control follows the strategy:
 $$
    u(t) = \left\{
    \begin{array}{lll}
        -1            & \text{if} & t \in [0, t_1],     \\[0.5em]
        \phantom{-}0  & \text{if} & t \in (t_1, t_2],   \\[0.5em]
        +1            & \text{if} & t \in (t_2, t_f],
    \end{array}
    \right.
 $$
 with $0 < t_1 < t_2 < t_f=2$.
 <div class="alert alert-warning">
 **Goal**
 The goal is to find the values of the switching times $t_1$ and $t_2$ together with the initial covector $p_0$ (see Remark 2).
 </div>
 %% Cell type:markdown id: tags:
 ### Maximized Hamiltonian and its derivatives
 %% Cell type:markdown id: tags:
 We define first the three control laws $u \equiv \{-1, 0, 1\}$.
 %% Cell type:code id: tags:
 ``` python
 # Controls in feedback form
 @tools.vectorize(vvars=(1, 2, 3))
 def uplus(t, x, p):
    u = +1.0
    return u
 @tools.vectorize(vvars=(1, 2, 3))
 def uminus(t, x, p):
    u = -1.0
    return u
 @tools.vectorize(vvars=(1, 2, 3))
 def using(t, x, p):
    u = 0.0
    return u
 ```
 %% Cell type:markdown id: tags:
 The pseudo-Hamiltonian is
 $$
    H(x,p,u) = pu - x^2.
 $$
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 4:_**
 Complete the code of the Hamiltonian for $u \equiv +1$, with its derivatives.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 4 to complete here
 # ----------------------------
 #
 # Definition of the Hamiltonian and its derivatives for u = 1
 #
 def dhfunplus(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
    hd = 0 ### TO COMPLETE: use uplus
    return hd
 def d2hfunplus(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    hdd = 0 ### TO COMPLETE
    return hdd
 @tools.tensorize(dhfunplus, d2hfunplus, tvars=(2, 3))
 def hfunplus(t, x, p):
    h = 0 ### TO COMPLETE: use uplus
    return h
 hplus = ocp.Hamiltonian(hfunplus)
 fplus = ocp.Flow(hplus)
 ```
 %% Cell type:markdown id: tags:
 We give the Hamiltonians for $u=-1$ and $u=0$ with their derivatives.
 %% Cell type:code id: tags:
 ``` python
 # Definition of the Hamiltonian and its derivatives for u = -1
 def dhfunminus(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
    u  = uminus(t, x, p)
    hd = u*dp - 2.0*x*dx
    return hd
 def d2hfunminus(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    hdd    = -2.0 * d2x * dx
    return hdd
 @tools.tensorize(dhfunminus, d2hfunminus, tvars=(2, 3))
 def hfunminus(t, x, p):
    u = uminus(t, x, p)
    h = p*u - x**2
    return h
 hminus = ocp.Hamiltonian(hfunminus)
 fminus = ocp.Flow(hminus)
 ```
 %% Cell type:code id: tags:
 ``` python
 # Definition of the Hamiltonian and its derivatives for u = 0
 def dhfunsing(t, x, dx, p, dp):
    # dh = dh_x dx + dh_p dp
    u  = using(t, x, p)
    hd = u*dp - 2.0*x*dx
    return hd
 def d2hfunsing(t, x, dx, d2x, p, dp, d2p):
    # d2h = dh_xx dx d2x + dh_xp dp d2x + dh_px dx d2p + dh_pp dp d2p
    hdd    = -2.0 * d2x * dx
    return hdd
 @tools.tensorize(dhfunsing, d2hfunsing, tvars=(2, 3))
 def hfunsing(t, x, p):
    u = using(t, x, p)
    h = p*u - x**2
    return h
 hsing = ocp.Hamiltonian(hfunsing)
 fsing = ocp.Flow(hsing)
 ```
 %% Cell type:markdown id: tags:
 ### Shooting function
 %% Cell type:markdown id: tags:
 The multiple shooting function is given by
 $$
 S(p_0, t_1, t_2) :=
 \begin{pmatrix}
     x(t_1, t_0, x_0, p_0, u_-) \\
     p(t_1, t_0, x_0, p_0, u_-) \\
-     x(t_f, t_2, x_2, p_2, u_+)
+     x(t_f, t_2, x_2, p_2, u_+) - 1/2
 \end{pmatrix},
 $$
 where $z_2 := (x_2, p_2) = z(t_2, t_1, x_1, p_1, u_0)$, $z_1 := (x_1, p_1) = z(t_1, t_0, x_0, p_0, u_-)$ and where z(t, s, a, b, u) is the solution at time $t$ of the Hamiltonian system associated to the control u starting at time $s$ at the initial condition $z(s) = (a,b)$.
 We have introduced the notation $u_-$ for $u\equiv -1$, $u_0$ for $u\equiv 0$ and $u_+$ for $u\equiv +1$.
 **_Remark:_** We know that $(x_2, p_2)=(0,0)$.
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 5:_**
 Complete the code of the multiple shooting function.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 5 to complete here
 # ----------------------------
 #
 # Multiple shooting function
 #
 tf = tf_final # we set the final time to the value tf_final
 def shoot_multiple(y):
    p0 = y[0]
    t1 = y[1]
    t2 = y[2]
    s = np.zeros([3]) ### TO COMPLETE: use fminus, fsin, fplus, t0, t1, t2, tf, x0, xf_target
    return s
 ```
 %% Cell type:markdown id: tags:
 ### Resolution of the shooting function
 %% Cell type:markdown id: tags:
 <div class="alert alert-info">
 **_Question 6:_**
 Give initial guesses for the times $t_1$ and $t_2$ according to the solution of the regularized problem.
 </div>
 %% Cell type:code id: tags:
 ``` python
 # ----------------------------
 # Answer 6 to complete here
 # ----------------------------
 #
 # Initial guess for the Newton solver
 t1_guess = 0.0 # to update
 t2_guess = 0.0 # to update
 p0_guess = sol_path_tf.xf # from previous homotopy
 ```
 %% Cell type:code id: tags:
 ``` python
 # Resolution of the shooting function
 y_guess  = np.array([float(p0_guess), t1_guess, t2_guess])
 sol_nle_mul  = nt.nle.solve(shoot_multiple, y_guess)
 ```
 %% Cell type:code id: tags:
 ``` python
 # function to plot solution
 def plotSolutionBSB(p0, t1, t2, tf):
    N      = 20
    tspan1  = list(np.linspace(t0, t1, N+1))
    tspan2  = list(np.linspace(t1, t2, N+1))
    tspanf  = list(np.linspace(t2, tf, N+1))
    x1, p1 = fminus(t0, x0, p0, tspan1)  # on [ 0, t1]
    x2, p2 =  fsing(t1, x1[-1], p1[-1], tspan2)  # on [t1, t2]
    xf, pf =  fplus(t2, x2[-1], p2[-1], tspanf)  # on [t2, tf]
    u1     = uminus(tspan1, x1, p1)
    u2     =  using(tspan2, x2, p2)
    uf     =  uplus(tspanf, xf, pf)
    fig = plt.figure()
    ax  = fig.add_subplot(511); ax.plot(tspan1, x1); ax.plot(tspan2, x2); ax.plot(tspanf, xf);
    ax.set_xlabel('t'); ax.set_ylabel('$x$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(513);  ax.plot(tspan1, p1); ax.plot(tspan2, p2); ax.plot(tspanf, pf);
    ax.set_xlabel('t'); ax.set_ylabel('$p$'); ax.axhline(0, color='k')
    ax  = fig.add_subplot(515);  ax.plot(tspan1, u1); ax.plot(tspan2, u2); ax.plot(tspanf, uf);
    ax.set_xlabel('t'); ax.set_ylabel('$u$'); ax.axhline(0, color='k')
 ```
 %% Cell type:code id: tags:
 ``` python
 # plot solution
 p0 = sol_nle_mul.x[0]
 t1 = sol_nle_mul.x[1]
 t2 = sol_nle_mul.x[2]
 plotSolutionBSB(p0, t1, t2, tf)
 ```